# Generalized variance of the sum of N correlated random variables

I am trying to model the variance of a time series $$Y_n$$ which is the sum of $$n$$ observations of $$X_i$$. I've reviewed the other answers on CrossValidated; however, I haven't been able to apply those solutions to this problem where I have known distributions of $$X_i$$ and known $$\rho_k$$ values. This is the closest answer but doesn't account for $$\rho_k \neq 0$$ with $$k>1$$: Variance of average of $n$ correlated random variables.

$$Y_n = \sum_{i=1}^{n} X_i$$

We can assume that $$X_i \sim N(\mu,\sigma^2)$$ and that they have some autocorrelation such that

$$Cor(X_{i+1},X_{i})=\rho_1\\ Cor(X_{i+2},X_{i})=\rho_2\\ \vdots\\ Cor(X_{i+k},X_{i})=\rho_k$$

Traditionally, if $$X_i$$ are uncorrelated, the variance of $$Y_n$$ is simple:

$$Var(Y_n) = n\sigma^2$$

However, because they are correlated... it's make the $$Y_n$$ calculation more clumsy. I have derived a solution for a few cases (small $$n$$) where $$\rho_k=0$$ for $$k>2$$ below:

\begin{aligned} Var(Y_5) &= Var(X_1+X_2+X_3+X_4+X_5)\\ \\ Var(Y_5) &= 5\sigma^2 + 2Cov(X_2,X_1) + 2Cov(X_3,X_2) + 2Cov(X_3,X_1) + 2Cov(X_4,X_2) + 2Cov(X_4,X_3) + 2Cov(X_5,X_4) + 2Cov(X_5,X_3)\\ \\ Var(Y_5) &= 5\sigma^2 + 2\rho_1 + 2\rho_1 + 2\rho_2 + 2\rho_2 + 2\rho_1 + 2\rho_1 + 2\rho_2\\ Var(Y_5) &= 5\sigma^2 + 8\rho_1 + 6\rho_2 \end{aligned}

I'm having a hard time getting a generalized solution of $$Y_n$$ even if we use this case where $$\rho_k=0$$ for $$k>2$$. I think it should look something like above but can't figure out the form in term so $$n$$, $$\rho$$, and $$\sigma^2$$. I think it should look something like this

$$Var(Y_n) = n\sigma^2 + f(n)\rho_1 + g(n)\rho_2$$

Any ideas would be very helpful. I'm sadly a little far removed from pen and paper math. Thanks

• Taking another look at stats.stackexchange.com/questions/391740/… ... I think this may be close: $Var(Y_n) = n\sigma^2 + 2(n(n-1))/n*\rho_1\sigma^2 + 2(n(n-2))/n*\rho_2\sigma^2$ for the n=5 case, $Var(Y_5) = 5\sigma^2 + 2(5(4))/5*\rho_1\sigma^2 + 2(5(3))/5*\rho_2\sigma^2$ $------->$ $Var(Y_5) = 5\sigma^2 + 8\rho_1\sigma^2 + 6\rho_2\sigma^2$ Commented Mar 8, 2021 at 14:44

It looks like you are supposing the covariance matrix of $$(X_1,X_2,\ldots,X_N)$$ is

$$\Sigma = \sigma^2\pmatrix{1 & \rho_1 & \rho_2 & \cdots & \rho_{N-1} \\ \rho_1 & 1 & \rho_1 & \cdots & \rho_{N-2}\\ \rho_2 & \rho_1 & 1 & \cdots & \rho_{N-3}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \rho_{N-1} & \rho_{N-2} & \rho_{N-3} & \cdots & 1} = (\sigma^2\rho_{|i-j|})_{1\le i\le N,\, 1 \le j\le N}$$

where, for the convenience of that final formula, I have set $$\rho_0 = 1.$$

Consider $$Y_m = X_1 + X_2 + \cdots + X_m$$ and $$Y_n = X_1 + X_2 + \cdots + X_n$$ where $$1 \le m, n\le N.$$ Writing $$1_k = (1,1,\ldots,1,0,0,\ldots,0)^\prime$$ for the vector with $$k$$ initial ones ($$k=0, 1, \ldots, N$$ are the possible values of $$k$$), you can read the covariance directly off the matrix product as

$$\operatorname{Cov}(Y_m,Y_n) = 1_m^\prime \Sigma 1_n$$

because this (obviously, by the rules of matrix multiplication) is the sum of all entries in the $$m\times n$$ upper left block of $$\Sigma,$$ which is

$$\operatorname{Cov}(Y_m,Y_n)= \sum_{i=1}^m \sum_{j=1}^n \Sigma_{ij} = \sigma^2\sum_{i=1}^m \sum_{j=1}^n \rho_{|i-j|}.$$

By the standard formula for correlation in terms of covariances, the correlation matrix of $$(Y_1, Y_2, \ldots, Y_N)$$ therefore has entries

$$\operatorname{Cor}(Y_m,Y_n) = \frac{\operatorname{Cov}(Y_m,Y_n)}{\sqrt{\operatorname{Cov}(Y_m,Y_m)\operatorname{Cov}(Y_n,Y_n)}}.$$

Because all the factors of $$\sigma$$ will cancel in this ratio, you may ignore them in the computation, whence

$$\operatorname{Cor}(Y_m,Y_n) = \frac{\sum_{i=1}^m \sum_{j=1}^n \rho_{|i-j|}}{\sum_{i=1}^m \sum_{j=1}^m \rho_{|i-j|}\sum_{i=1}^n \sum_{j=1}^n \rho_{|i-j|}}.$$

These double sums can be expressed a little more simply, after reversing the roles of $$Y_m$$ and $$Y_n$$ if necessary to assure $$m\le n,$$ where

\begin{aligned} \operatorname{Cov}(Y_m,Y_n) &= \sigma^2 m(\rho_1+\rho_2+\cdots+\rho_{n-m}) + \sigma^2\sum_{j=1}^m (m-j)(\rho_j + \rho_{n-m+j}). \end{aligned}

Applying this to the case $$m=n$$ gives

$$\operatorname{Var}(Y_m) = \sigma^2\left(m + \sum_{i=1}^{m-1} 2(m-i)\rho_i\right).$$

For example,

$$\operatorname{Var}(Y_5) = \sigma^2(5 + 8\rho_1 + 6\rho_2 + 4\rho_3 + 2\rho_4).$$

• Thank you very much for the thorough answer. Makes sense! At least I was on the right track Commented Mar 8, 2021 at 21:11